Without loss of generality, we can work with 2-dimensional matrices.

The Abelian group of order 3 has these elements:

In[429]:= A3 = {IdentityMatrix[2], RotationMatrix[2 \[Pi]/3], 
  RotationMatrix[4 \[Pi]/3]}

Out[429]= {{{1, 0}, {0, 1}}, {{-(1/2), -(Sqrt[3]/2)}, {Sqrt[3]/
   2, -(1/2)}}, {{-(1/2), Sqrt[3]/2}, {-(Sqrt[3]/2), -(1/2)}}}

We can define a function to compute the order of a group element by iterating over matrix multiplication:

In[430]:= groupElementOrder[m_?MatrixQ] := 
 Module[{n = 1, e = IdentityMatrix[Length[m]]}, 
  NestWhile[(n++; Dot[#, m]) &, m, # != e &]; n]

NestWhile does the iteration, and terminates when the matrix product equals the identity matrix. Now, we can test it on the group A3:

In[431]:= groupElementOrder /@ A3

Out[431]= {1, 3, 3}

I am not sure exactly what you are asking, but perhaps this might give you an idea for a direction.

Here is the finite list generated by repeated multiplications by I:

FixedPoint[Union[Join[#, I #]] &, {1}, 1000]

So the order is:

Length[  FixedPoint[Union[Join[#, I #]] &, {1}, 1000]]

For multiplications by Exp[Pi I/12]:

FixedPoint[Union[Join[#, Exp[Pi I/12] #]] &, {1}, 10000]